航空器的稳定与控制（英文版）：lecture_4

16.333 Lecture 4 Aircraft Dynamics Aircraft nonlinear EOM ? ? Linearization – dynamics Linearization – forces & moments ? ? Stability derivatives and coe?cients Fall 2004 16.333 4–1 Aircraft Dynamics ? Note can develop good approximation of key aircraft motion (Phugoid) using simple balance between kinetic and potential energies. ? Consider an aircraft in steady, level ?ight with speed U 0 and height h 0 . The motion is perturbed slightly so that U 0 U = U 0 + u (1)→ h 0 h = h 0 + Δh (2)→ ? Assume that E = 1 mU 2 + mgh is constant before and after the 2 perturbation. It then follows that u ≈? gΔh U 0 ? From Newton’s laws we know that, in the vertical direction ¨ mh = L ?W 1 where weight W = mg and lift L = 2 ρSC L U 2 (S is the wing area). We can then derive the equations of motion of the aircraft: ¨ 1 mh = L ?W = ρSC L (U 2 ?U 0 2 ) (3) 2 1 = ρSC L ((U 0 + u) 2 ?U 0 2 ) ≈ 1 ρSC L (2uU 0 )(4) 22 ? ? gΔh U 0 = ?(ρSC L g)Δh (5)≈ ?ρSC L U 0 ¨ ¨ Since h = Δh and for the original equilibrium ?ight condition L = 1 W = 2 (ρSC L )U 2 = mg, we get that 0 ? ? 2 ρSC L g g = 2 m U 0 Combine these result to obtain: ¨ Δh + Ω 2 Δh = 0 , Ω ≈ g √ 2 U 0 ? These equations describe an oscillation (called the phugoid oscilla- tion) of the altitude of the aircraft about it nominal value. – Only approximate natural frequency (Lanchester), but value close. ? k) ? ? ? Fall 2004 16.333 4–2 ? The basic dynamics are: ? ˙ I ? ˙ I F = mv c and T ? = H 1 ? ˙ B ω × ?v c Transport Thm. F = v c + BI ?? m ? ˙ B BI ? ? T ? = H + ω × ? H ? Basic assumptions are: 1. Earth is an inertial reference frame 2. A/C is a rigid body 3. Body frame B ?xed to the aircraft ( ? i, ? j, BI ?? Instantaneous mapping of ?v c and ω into the body frame: BI ?ω = P ? i + Q ? j + R ? k ?v c = U ? i + V ? j + W ? k ? ? ? ? P U ? ? BI ω B = ? Q ? ? (v c ) B = ? V R W ? By symmetry, we can show that I xy = I yz = 0, but value of I xz depends on speci?c frame selected. Instantaneous mapping of the angular momentum H = H x ? i + H y ? j + H z ? k into the Body Frame given by ? ? ? ?? ? H x I xx 0 I xz P H B = ? H y ? = ? 0 I yy 0 ?? Q ? H z I xz 0 I zz R ? 1 Fall 2004 16.333 4–3 ? The overall equations of motion are then: 1 ? ˙ B F = v c + BI ?ω × ?v c m ? ? ? ? ? ?? ? X U ˙ 0 ?R Q U ? m ? Y ? = ? V ˙ ? + ? 0 ?P ?? V ? R ˙ Z W ?Q P 0 W ? ? U ˙ +QW ? RV ˙ ? = ? V +RU ? P W ˙ W +P V ? QU ? ˙ B BI ?T ? = H + ω × ? H ? ? ? ? ? ?? ?? ? ˙ L I xx P ˙ +I xz R 0 ?R Q I xx 0 I xz P ? ? M ? = ? ˙ I yy Q ˙ ? + ? 0 ?P ?? 0 I yy 0 ?? Q ? R P 0 I xz 0 I zz RN I zz R +I xz P ˙ ?Q ? ? ˙ I xx P ˙ +I xz R +QR(I zz ? I yy ) +P QI xz ? = ? I yy Q ˙ +P R(I xx ? I zz ) + (R 2 ? P 2 )I xz ˙ I zz R +I xz P ˙ +P Q(I yy ? I xx )? QRI xz ? Clearly these equations are very nonlinear and complicated, and we have not even said where ? F and T ? come from. = Need to linearize!! ? – Assume that the aircraft is ?ying in an equilibrium condition and we will linearize the equations about this nominal ?ight condition. Fall 2004 16.333 4–4 Axes ? But ?rst we need to be a little more speci?c about which Body Frame we are going use. Several standards: 1. Body Axes - X aligned with fuselage nose. Z perpendicular to X in plane of symmetry (down). Y perpendicular to XZ plane, to the right. 2. Wind Axes - X aligned with ?v c . Z perpendicular to X (pointed down). Y perpendicular to XZ plane, o? to the right. 3. Stability Axes - X aligned with projection of ?v c into the fuselage plane of symmetry. Z perpendicular to X (pointed down). Y same. R E L A T I V E W I N D ( ) ( ) ( ) ? ? B ODY Z-AXIS B ODY Y -AXIS X-AXIS WIND X-AXIS S T AB ILIT Y X-AXIS B ODY ? Advantages to each, but typically use the stability axes. – In di?erent ?ight equilibrium conditions, the axes will be oriented di?erently with respect to the A/C principal axes ? need to trans- form (rotate) the principal inertia components between the frames. – When vehicle undergoes motion with respect to the equilibrium, Stability Axes remain ?xed to airplane as if painted on. Fall 2004 16.333 4–5 ? Can linearize about various steady state conditions of ?ight. – For steady state ?ight conditions must have F F aero + ? F thrust = 0 and T ? = 0 ? = ? F gravity + ? 3 So for equilibrium condition, forces balance on the aircraft L = W and T = D ˙ ˙ ˙ ˙ ˙ – Also assume that P ˙ = Q = R = U = V = W = 0 – Impose additional constraints that depend on ?ight condition: 3 Steady wings-level ?ight → Φ = ˙ Θ = ˙ Φ = ˙ Ψ = 0 ? Key Point: While nominal forces and moments balance to zero, motion about the equilibrium condition results in perturbations to the forces/moments. – Recall from basic ?ight dynamics that lift L f = C L α α 0 where: 0 3 C L α = lift curve slope – function of the equilibrium condition 3 α 0 = nominal angle of attack (angle that wing meets air ?ow) – But, as the vehicle moves about the equilibrium condition, would expect that the angle of attack will change α = α 0 + Δα – Thus the lift forces will also be perturbed L f = C L α (α 0 + Δα) = L f + ΔL f 0 ? Can extend this idea to all dynamic variables and how they in?uence all aerodynamic forces and moments Fall 2004 16.333 4–6 Gravity Forces ? Gravity acts through the CoM in vertical direction (inertial frame +Z) – Assume that we have a non-zero pitch angle Θ 0 – Need to map this force into the body frame – Use the Euler angle transformation (2–15) ? ? ? ? 0 ? sin Θ F g = T 1 (Φ)T 2 (Θ)T 3 (Ψ) ? 0 ? = mg ? sin Φ cos Θ ? B mg cos Φ cos Θ ? For symmetric steady state ?ight equilibrium, we will typically assume that Θ ≡ Θ 0 , Φ ≡ Φ 0 = 0, so ? ? ? sin Θ 0 F g = mg ? 0 ? B cos Θ 0 ? Use Euler angles to specify vehicle rotations with respect to the Earth frame Θ ˙ = Q cos Φ ? R sin Φ Φ ˙ = P + Q sin Φ tan Θ + R cos Φ tan Θ Ψ ˙ = (Q sin Φ + R cos Φ) sec Θ – Note that if Φ ≈ 0, then Θ ˙ ≈ Q ? Recall: Φ ≈ Roll, Θ ≈ Pitch, and Ψ ≈ Heading. Fall 2004 16.333 4–7 Linearization ? De?ne the trim angular rates and velocities ? ? ? ? P U o BI ω o = ? Q ? (v c ) o = ? 0 ? B B R 0 which are associated with the ?ight condition. In fact, these de?ne the type of equilibrium motion that we linearize about. Note: – W 0 = 0 since we are using the stability axes, and – V 0 = 0 because we are assuming symmetric ?ight ? Proceed with linearization of the dynamics for various ?ight conditions Nominal Perturbed Perturbed ? Velocity Velocity Acceleration? Velocities U 0 , U = U 0 +u U ˙ = u˙? ˙ W 0 = 0, W = w W = w˙ V 0 = 0, V = v ? V ˙ = v˙? Angular P 0 = 0, Rates Q 0 = 0, R 0 = 0, P = p P ˙ = ˙p Q = q ? Q ˙ = ˙q R = r ? R ˙ = r˙? ˙ Angles Θ 0 , Θ = Θ 0 +θ Θ = θ ˙ ? ˙ Φ 0 = 0, Φ = φ Φ = φ ˙ ? ˙ Ψ 0 = 0, Ψ = ψ Ψ = ψ ˙ ? ? Fall 2004 16.333 4–8 W C.G. 0 V T V T 0 = U 0 X E X 0 X q ? Z Z 0 Z E ? Θ ? or ? 0 ? 0 ? 0 Horizontal U = U + u Down Figure 1: Perturbed Axes. The equilibrium condition was that the aircraft was angled up by Θ 0 with velocity V T 0 = U 0 . The vehicle’s motion has been perturbed (X 0 → X) so that now Θ = Θ 0 +θ and the velocity is V T = V T 0 . Note that V T is no longer aligned with the X-axis, resulting in a non-zero u and w. The angle γ is called the ?ight path angle, and it provides a measure of the angle of the velocity vector to the inertial horizontal axis. Fall 2004 16.333 4–9 ? Linearization for symmetric ?ight U = U 0 +u, V 0 = W 0 = 0, P 0 = Q 0 = R 0 = 0. Note that the forces and moments are also perturbed. 1 [X 0 + ΔX] = U ˙ +QW ? RV u˙ +qw ? rv ≈ u˙ m ≈ 1 [Y 0 + ΔY ] = V ˙ +RU ? P W m v˙ +r(U 0 +u)? pw v˙ +rU 0 ≈ ≈ 1 ˙ [Z 0 + ΔZ] = W +P V ? QU w˙ +pv ? q(U 0 +u) m ≈ w˙ ? qU 0 ≈ ? ? ? ? ΔX u˙ 1 1 ? ΔY ? = ? v˙ +rU 0 ? 2? m ΔZ w˙ ? qU 0 3 Attitude motion: ? ? ? ? ? ˙ L I xx P ˙ +I xz R +QR(I zz ? I yy ) +P QI xz ?? M ? = ? I yy Q ˙ +P R(I xx ? I zz ) + (R 2 ? P 2 )I xz ˙ N I zz R +I xz P ˙ +P Q(I yy ? I xx )? QRI xz ? ? ? ? ΔL I xx p˙ +I xz r˙ 4 ? ΔM ? = ? I yy q˙ ? 5 ? ΔN I zz r˙ +I xz p˙ 6 Fall 2004 16.333 4–10 ? Key aerodynamic parameters are also perturbed: Total Velocity 2 2 ) 1/2 ≈ U 0 + uV T = ((U 0 + u) 2 + v + w Perturbed Sideslip angle β = sin ?1 (v/V T ) ≈ v/U 0 Perturbed Angle of Attack α x = tan ?1 (w/U) ≈ w/U 0 ? To understand these equations in detail, and the resulting impact on the vehicle dynamics, we must investigate the terms ΔX . . . ΔN. – We must also address the left-hand side ( ? F, T ? ) – Net forces and moments must be zero in equilibrium condition. – Aerodynamic and Gravity forces are a function of equilibrium con- dition AND the perturbations about this equilibrium. ? Predict the changes to the aerodynamic forces and moments using a ?rst order expansion in the key ?ight parameters ?X ?X ?X ?X ?X g ΔX = ΔU + ΔW + Δ ˙ W + ΔΘ + . . . + ΔΘ + ΔX c ?U ?W ? ˙ W ?Θ ?Θ ?X ?X ?X ?X ?X g ˙= u + w + w + θ + . . . + θ + ΔX c ?U ?W ? ˙ W ?Θ ?Θ ? ?X called stability derivative – evaluated at eq. condition. ?U ? Gives dimensional form; non-dimensional form available in tables. ? Clearly approximation since ignores lags in the aerodynamics forces (assumes that forces only function of instantaneous values) Fall 2004 16.333 4–11 Stability Derivatives ? First proposed by Bryan (1911) – has proven to be a very e?ec- tive way to analyze the aircraft ?ight mechanics – well supported by numerous ?ight test comparisons. ? The forces and torques acting on the aircraft are very complex nonlin- ear functions of the ?ight equilibrium condition and the perturbations from equilibrium. – Linearized expansion can involve many terms u, ˙ u, . . . , w, ˙ w, . . . u, ¨ w, ¨ – Typically only retain a few terms to capture the dominant e?ects. ? Dominant behavior most easily discussed in terms of the: – Symmetric variables: U, W, Q & forces/torques: X, Z, and M – Asymmetric variables: V , P, R & forces/torques: Y , L, and N ? Observation – for truly symmetric ?ight Y , L, and N will be exactly zero for any value of U, W , Q ? Derivatives of asymmetric forces/torques with respect to the sym- metric motion variables are zero. ? Further (convenient) assumptions: 1. Derivatives of symmetric forces/torques with respect to the asym- metric motion variables are small and can be neglected. 2. We can neglect derivatives with respect to the derivatives of the motion variables, but keep ?Z/?w˙ and M w˙ ≡ ?M/?w˙ (aero- dynamic lag involved in forming new pressure distribution on the wing in response to the perturbed angle of attack) 3. ?X/?q is negligibly small. ? ? ? ? ? ? ? ? ? ? ? ? Fall 2004 16.333 4–12 ?()/?() X Y Z L M N u v w p q r ? 0 ? 0 ≈ 0 0 0 ? 0 ? 0 ? ? 0 ? 0 ? 0 0 ? 0 ? 0 ? ? 0 ? 0 ? 0 0 ? 0 ? 0 ? ? Note that we must also ?nd the perturbation gravity and thrust forces and moments ?X g ?Θ ?Z g = ?mg cos Θ 0 0 ?Θ = ?mg sin Θ 0 0 ? Aerodynamic summary: 1A ΔX = ?X ?U 0 u + ?X ?W 0 w ? ΔX ～ u, α x ≈ w/U 0 2A ΔY ～ β ≈ v/U 0 , p, r 3A ΔZ ～ u, α x ≈ w/U 0 , ˙α x ≈ ˙w/U 0 , q 4A ΔL ～ β ≈ v/U 0 , p, r 5A ΔM ～ u, α x ≈ w/U 0 , ˙α x ≈ ˙w/U 0 , q 6A ΔN ～ β ≈ v/U 0 , p, r ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Fall 2004 16.333 4–13 ? Result is that, with these force, torque approximations, equations 1, 3, 5 decouple from 2, 4, 6 – 1, 3, 5 are the longitudinal dynamics in u, w, and q ???? ΔX mu˙ ΔZ ? = ? m( ˙w ? qU 0 ) ? ΔM I yy q˙ ?? ?X g ?X ?X θ + ΔX c u + w + ?U ?W ?Θ0 0 0 ? ? ? ? ? ? ?Z g ?Z ?Z ?Z ?Z w + θ + ΔZ c ˙u + w + q + ≈ ? ˙ W ?U ?W ?Q ?Θ0 0 0 0 0 ?M ?M ?M w˙ + ?M q + ΔM c u + w + ?U 0 ?W 0 ? ˙ W 0 ?Q 0 – 2, 4, 6 are the lateral dynamics in v, p, and r ???? ΔY m(˙v +rU 0 ) ? ΔL ? = ? I xx p˙ +I xz r˙ ? ΔN I zz r˙ +I xz p˙ ?? ≈ ? ? ? ? ? ?Y ?Y ?Y v + 0 p + r + +ΔY c ?V 0 ?P ?R 0 ?L v + ?L 0 p + ?L r + ΔL c ?V 0 ?P ?R 0 ? ? ? ? ? ? ? ? ? ? ? ? ?N ?N ?N v + 0 p + r + ΔN c ?V 0 ?P ?R 0 ? ? ? ? ? ? ? ? ? ? ? ? Fall 2004 16.333 4–14 Basic Stability Derivative Derivation ? Consider changes in the drag force with forward speed U D = 1/2ρV T 2 SC D 2 2 V 2 = (u 0 +u) 2 +v +w T ?V T 2 ?V T 2 = 2(u 0 +u) ? = 2u 0 ?u ?u 0 ?V T 2 ?V T 2 Note: = 0 and = 0 ?v 0 ?w 0 At reference condition: ? ? ? ? ?? ?D ? ρV T 2 SC D = ? D u ≡ ?u 0 ?u 2 ? 0 ? ? ? ?? ρS 2 ?C D ?V T 2 = u 0 +C D 0 2 ?u 0 ?u 0 ρS 2 ?C D = + 2u 0 C D 0 u 0 2 ?u 0 – Note ?D is the stability derivative, which is dimensional. ?u ? De?ne nondimensional stability coe?cient C D u as derivative of C D with respect to a nondimensional velocity u/u 0 D ?C D and C D 0 ≡ (C D ) 0 C D = 1 ρV T 2 S ? C D u ≡ ?u/u 0 2 0 – So ( 0 corresponds to the variable at its equilibrium condition. ?) ?? ? ? Fall 2004 16.333 4–15 Nondimensionalize: ? ? ? ? ? ? ? ?D ρSu 0 ?C D = u 0 + 2C D 0 ?u 0 2 ?u 0 QS ?C D = + 2C D 0 u 0 ?u/u 0 0 ? ? ? ? u 0 ?D = (C D u + 2C D 0 ) QS ?u 0 So given stability coe?cient, can compute the drag force increment. ? Note that Mach number has a signi?cant e?ect on the drag: ? ? ? ? C D u = ?C D ?u/u 0 0 = u 0 a ?C D ? ? u a ? 0 = M ?C D ?M where ?C D ?M can be estimated from empirical results/tables. Aerodynamic Principles ? = Constant C D 0 M cr 0 1 2 = 0.1 C D M Mach number, M ? Thrust forces ? ? ? ? C T u = ?C T ?u/u 0 0 ? ?T ?u 0 = C T u 1 u 0 QS – For a glider, C T u = 0 – For a jet, C T u ≈ 0 – For a prop plane, C T u = ?C D 0 ?? ? Fall 2004 16.333 4–16 ? Lift forces similar to drag L = 1/2ρV T 2 SC L ? ? ? ? ? ? ?L ρSu 0 ?C L = + 2C L 0 ? ?u 0 2 u 0 ?u 0 ? QS ?C L = + 2C L 0 u 0 ?u/u 0 0 ? ? ? ? u 0 ?L = (C L u + 2C L 0 ) QS ?u 0 where C L 0 is the lift coe?cient for the eq. condition and C L u = M ?C L ?M as before. From aerodynamic theory, we have that ?C L MC L | M=0 C L = = C L √ 1? M 2 ? ?M 1? M 2 M 2 ? C L u = 1? M 2 C L 0 ? α Derivatives: Now consider what happens with changes in the angle of attack. Take derivatives and evaluate at the reference condition: – Lift: ? C L α C 2 L 2C L 0 – Drag: C D = C D min + πeAR ? C D α = πeAR C L α ? ? ? ? Fall 2004 16.333 4–17 ? Combine into X, Z Forces – At equilibrium, forces balance. – Use stability axes, so α 0 = 0 – Include the e?ect in the force balance of a change in α on the force rotations so that we can see the perturbations. – Assume perturbation α is small, so rotations are by cos α ≈ 1, sin α ≈ α X = T ? D + Lα Z = ?(L + Dα) C L C x C z z x V Note: ? = ? T at t = 0 C D ??(t) (i.e., Static Trim) ? So, now consider the α derivatives of these forces: ?X ?T ?D ?L = + L + α ?α ?α ? ?α ?α ?T – Thrust variation with α very small ?α 0 ≈ 0 – Apply at the reference condition (α = 0), i.e. C X α ? Nondimensionalize and apply reference condition: C X α = ?C D α + C L 0 2C L 0 = C L 0 ? C L α πeAR ?C X = ?α 0 Fall 2004 16.333 4–18 And for the Z direction ? ?Z ?D ?L = D ?α ?α ? ?α ? ?α Giving C Z α = C D 0 ?C L α ? ? Recall that C M α was already found during the static analysis ? Can repeat this process for the other derivatives with respect to the forward speed. ? Forward speed: ?X ?T ?D ?L = +α ?u ?u ? ?u ?u So that ? ?? ? ? ?? ? ? ?? ? u 0 ?X u 0 ?T u 0 ?D = QS ?u 0 QS ?u 0 ? QS ?u 0 ?C X u ≡ C T u ?(C D u + 2C D 0 ) ? Similarly for the Z direction: ?Z ?L ?D = α ?u ? ?u ? ?u So that ? ?? ? ? ?? ? u 0 ?Z u 0 ?L = QS ?u 0 ? QS ?u 0 (C L u + 2C L 0 )C Z u ≡ ? M 2 = ? 1?M 2 C L 0 ?2C L 0 ? Many more derivatives to consider ! Fall 2004 16.333 4–19 Summary ? Picked a speci?c Body Frame (stability axes) from the list of alter- natives ? Choice simpli?es some of the linearization, but the inertias now change depending on the equilibrium ?ight condition. ? Since the nonlinear behavior is too di?cult to analyze, we needed to consider the linearized dynamic behavior around a speci?c ?ight condition ? Enables us to linearize RHS of equations of motion. ? Forces and moments also complicated nonlinear functions, so we lin- earized the LHS as well ? Enables us to write the perturbations of the forces and moments in terms of the motion variables. – Engineering insight allows us to argue that many of the stability derivatives that couple the longitudinal (symmetric) and lateral (asymmetric) motions are small and can be ignored. ? Approach requires that you have the stability derivatives. – These can be measured or calculated from the aircraft plan form and basic aerodynamic data.